Largest BST.cpp

/*
Largest BST
===========

Given a binary tree. Find the size of its largest subtree that is a Binary Search Tree.

Example 1:
Input:
        1
      /   \
     4     4
   /   \
  6     8
Output: 1
Explanation: There's no sub-tree with size
greater than 1 which forms a BST. All the
leaf Nodes are the BSTs with size equal
to 1.

Example 2:
Input: 6 6 3 N 2 9 3 N 8 8 2
            6
        /       \
       6         3
        \      /   \
         2    9     3
          \  /  \
          8 8    2 
Output: 2
Explanation: The following sub-tree is a
BST of size 2: 
       2
    /    \ 
   N      8
Your Task:
You don't need to read input or print anything. Your task is to complete the function largestBst() that takes the root node of the Binary Tree as its input and returns the size of the largest subtree which is also the BST. If the complete Binary Tree is a BST, return the size of the complete Binary Tree. 

Expected Time Complexity: O(N).
Expected Auxiliary Space: O(Height of the BST).

Constraints:
1 <= Number of nodes <= 100000
1 <= Data of a node <= 1000000
*/

struct NodeCheck
{
  bool isBST;
  int minNode, maxNode;
  int noOfNodes;
};

NodeCheck dfs(Node *root, int &ans)
{
  if (!root)
  {
    NodeCheck sub_ans;
    sub_ans.isBST = true;
    sub_ans.minNode = INT_MAX;
    sub_ans.maxNode = INT_MIN;
    sub_ans.noOfNodes = 0;
    return sub_ans;
  }

  auto left = dfs(root->left, ans);
  auto right = dfs(root->right, ans);

  int curr_count = 0;
  bool curr_bst = false;

  if (left.isBST && right.isBST && left.maxNode < root->data && right.minNode > root->data)
  {
    curr_count = 1 + left.noOfNodes + right.noOfNodes;
    curr_bst = true;
  }

  ans = max(ans, curr_count);

  NodeCheck sub_ans;
  sub_ans.isBST = curr_bst;
  sub_ans.minNode = min(left.minNode, root->data);
  sub_ans.maxNode = max(right.maxNode, root->data);
  sub_ans.noOfNodes = curr_count;

  return sub_ans;
}

int largestBst(Node *root)
{
  int ans = 0;
  dfs(root, ans);
  return ans;
}